Download Tutorial on the normal distribution

The Normal Distribution
December 9, 2016
Contents
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Normal Distribution Tutorial
Converting scores to areas
Converting areas to scores
Questions
Normal Distribution Tutorial
Normal distributions can only vary by their means and standard distributions. If you know a distribution is normal, and you
know the mean and standard deviation, then you have everything you need to know to calculate areas and probabilities.
To find areas under any normal distribution we convert our scores into z-scores and look up the answer in the z-table. Given
a normal distribution of scores, X, that has a mean µ and standard deviation σ, we convert X to z with:
z=
X −µ
σ
Example: Suppose you know that a population of test scores has a mean of 70 and a standard deviation of 10. What
proportion of scores fall above 80?
These problems can be solved by converting scores into standard deviation units and then looking up values in the z-table.
To convert scores to standard deviation units, we subtract the mean and then divide by the standard deviation:
z=
X−µ
80−70
σ = 10 = 1
The standard normal distribution tells you the proportion of areas under the normal distribution in standard deviation units.
So the proportion of scores above X = 80 is the same as the area under the standard normal distribution above z = 1:
area =0.1587
80
40
50
60
70
Scores (X)
1
80
90
100
area =0.1587
1
-3
-2
-1
0
1
2
3
z
From the z-table, this area is 0.1587.
Converting scores to areas
Example: What porportion of scores fall either below 55 or above 90?
This time we need to convert two test scores into z-scores:
z1 =
X1 −µ
= 55−70
σ
10 = −1.5
z2 =
X2 −µ
= 90−70
σ
10 = 2
area =0.0668
area =0.0228
55
40
50
90
60
70
Scores (X)
2
80
90
100
area =0.0668
area =0.0228
-1.5
-3
-2
2
-1
0
1
2
3
z
We then use the z-table to find the two areas and add them up: 0.0668 + 0.0228 = 0.0896
Converting areas to scores
If we’re given an area under the standard normal distribution and we need to convert it to a score, X, we first find the z-score
for that area in the z-table and use:
X = zσ + µ
Example: For what test score does 5% of the scores lie above?
To go from areas to scores we first find the z-score for the area. Using the z-table we find that the z-score for which 5% of
the area lies above is z = 1.64. To convert z-scores to test scores we multiply z by the population standard deviation and
add the mean:
X = (z)(σ) + µ = (1.64)(10) + 70 = 86.45
area =0.05
1.64
-3
-2
-1
0
z
3
1
2
3
area =0.05
86.45
40
50
60
70
80
90
100
Scores (X)
Example: What test scores bracket the middle 95% of all scores?
area =0.025
area =0.025
-1.96
-3
-2
1.96
-1
0
z
4
1
2
3
area =0.025
area =0.025
50.4
40
50
89.6
60
70
80
90
100
Scores (X)
We first find the z-scores that bracket the middle 95% of all z scores. This corresponds to finding the z-score for which
(100-95)/2 = 2.5% lies above and below. Using the table, we find that z = -1.96 and z = 1.96.
We then convert these two z-scores to test scores:
X1 = (z1 )(σ) + µ = (−1.96)(10) + 70 = 50.4
X2 = (z2 )(σ) + µ = (1.96)(10) + 70 = 89.6
So 95% of all test scores fall between 50.4 and 89.6.
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Questions
Now it’s your turn. Here are 28 problems with answers. For all examples, assume that the population is normally distributed.
Draw pictures if it helps!
1) Suppose the peace of brains has a mean of 22 and a standard deviation of 6.2.
Find the value for which 26.43 percent of the peace falls above.
Answer: 25.90
2) Suppose the arousal of candy bars has a mean of 92 and a standard deviation of 5.1.
Find the proportion of the arousal below 88.4 and above 93.8
Answer: 0.6051
3) Suppose the conduct of psychologists has a mean of 95 and a standard deviation of 6.7.
Find the proportion of the conduct below 95.4 and above 107.1
Answer: 0.5599
4) Suppose the amount of body parts has a mean of 76 and a standard deviation of 5.3.
Find the proportion of the amount between 68.3 and 80.6
Answer: 0.7343
5) Suppose the scenery of fingers has a mean of 33 and a standard deviation of 4.3.
Find the proportion of the scenery between 30.2 and 38.4
Answer: 0.6333
6) Suppose the peace of fingerprints has a mean of 38 and a standard deviation of 5.6.
Find the value for which 64.06 percent of the peace falls above.
Answer: 36.00
7) Suppose the equipment of Seattleites has a mean of 4 and a standard deviation of 4.9.
Find values that bracket the middle 82.30 percent of the equipment.
Answer: between -2.60 and 10.60
8) Suppose the sadness of geeks has a mean of 3 and a standard deviation of 3.7.
Find values that bracket the middle 67.29 percent of the sadness.
Answer: between -0.60 and 6.60
9) Suppose the shopping of airlines has a mean of 65 and a standard deviation of 5.
Find the proportion of the shopping between 61 and 61.2
Answer: 0.0088
10) Suppose the melancholy of antidepressants has a mean of 1 and a standard deviation of 5.9.
Find the proportion of the melancholy below 0.6 and above 4
Answer: 0.7811
11) Suppose the body mass index of planets has a mean of 32 and a standard deviation of 5.4.
Find the value for which 22.36 percent of the body mass index falls above.
Answer: 36.10
12) Suppose the mail of cats has a mean of 64 and a standard deviation of 5.4.
Find the proportion of the mail below 56.8 and above 66.1
Answer: 0.4421
13) Suppose the smell of response times has a mean of 2 and a standard deviation of 6.1.
Find the proportion of the smell between -12.5 and 6.5
Answer: 0.7586
14) Suppose the grief of psychology classes has a mean of 63 and a standard deviation of 4.7.
Find the value for which 62.55 percent of the grief falls above.
Answer: 61.50
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15) Suppose the piety of infants has a mean of 51 and a standard deviation of 5.3.
Find the value for which 69.50 percent of the piety falls above.
Answer: 48.30
16) Suppose the sadness of diseases has a mean of 76 and a standard deviation of 5.
Find the proportion of the sadness above 84.1
Answer: 0.0537
17) Suppose the distance of Europeans has a mean of 53 and a standard deviation of 5.8.
Find values that bracket the middle 21.28 percent of the distance.
Answer: between 51.40 and 54.60
18) Suppose the duration of chickens has a mean of 18 and a standard deviation of 4.1.
Find the proportion of the duration above 17.7
Answer: 0.5279
19) Suppose the recognition of cell phones has a mean of 10 and a standard deviation of 5.1.
Find the value for which 71.23 percent of the recognition falls above.
Answer: 7.10
20) Suppose the advice of cartoon characters has a mean of 40 and a standard deviation of 4.7.
Find values that bracket the middle 32.55 percent of the advice.
Answer: between 38.00 and 42.00
21) Suppose the progress of republicans has a mean of 45 and a standard deviation of 6.6.
Find the proportion of the progress between 28.9 and 37.7
Answer: 0.1283
22) Suppose the cash of PhDs has a mean of 49 and a standard deviation of 6.1.
Find value for which 71.90 percent of the cash fall below.
Answer: 52.50
23) Suppose the justice of chickens has a mean of 28 and a standard deviation of 4.7.
Find values that bracket the middle 31.82 percent of the justice.
Answer: between 26.10 and 29.90
24) Suppose the scenery of elbows has a mean of 28 and a standard deviation of 4.
Find the value for which 84.61 percent of the scenery falls above.
Answer: 23.90
25) Suppose the importance of airlines has a mean of 27 and a standard deviation of 5.3.
Find the proportion of the importance above 35.1
Answer: 0.0643
26) Suppose the information of psych 315 students has a mean of 83 and a standard deviation of 5.7.
Find value for which 30.50 percent of the information fall below.
Answer: 80.10
27) Suppose the distance of colors has a mean of 59 and a standard deviation of 5.9.
Find the proportion of the distance between 54.7 and 57.5
Answer: 0.1686
28) Suppose the anger of skittles has a mean of 30 and a standard deviation of 4.6.
Find the value for which 72.91 percent of the anger falls above.
Answer: 27.20
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